Question: Simplify; express your answer in exponential form. Assume $p\neq 0, n\neq 0$. $\dfrac{{p^{-1}}}{{(p^{2}n^{4})^{3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${p^{-1}}$ to the exponent ${1}$ . Now ${-1 \times 1 = -1}$ , so ${p^{-1} = p^{-1}}$ In the denominator, we can use the distributive property of exponents. ${(p^{2}n^{4})^{3} = (p^{2})^{3}(n^{4})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{p^{-1}}}{{(p^{2}n^{4})^{3}}} = \dfrac{{p^{-1}}}{{p^{6}n^{12}}}$ Break up the equation by variable and simplify. $\dfrac{{p^{-1}}}{{p^{6}n^{12}}} = \dfrac{{p^{-1}}}{{p^{6}}} \cdot \dfrac{{1}}{{n^{12}}} = p^{{-1} - {6}} \cdot n^{- {12}} = p^{-7}n^{-12}$.